Question

what is dimensional formula of permitivity of free space or vaccum? plz ans fast plz....

Answer 1

HEY MATE

HERE IS UR A NS

charge ×charge /force distance

[M-1L-3T4A2]

HOPE IT HELPS U ^_^...

Answer 2

we know that Electrostatic force ( coulomb ' s force )
F =( K q1 × q2 ) / r ^2

=> k = ( F × r^2 ) / q1 × q2 ---- ( i )
here k = 1 / ( 4 × Pi × € ) Putting in ( i ) gives
1 / ( 4 × Pi × € ) = ( F × r^2 ) / q1 × q2
taking reciprocal on both sides
4 × Pi × € = ( q1 × q2 ) / ( F × r^2 )
=> € = ( q1 × q2 ) / (4 × Pi × F × r^2 ) ---(ii)
where € = permitivity of free space

Now
q1 and q2 are charges and SI unit charge is Coulomb denoted by symbol C...
We know that current , I = q / t
=> q = I × t => C = [ A ] × [T]
where I = current , its SI unit is Ampere denoted by A and its dimesion is [A]
t = time , Dimension of time = [T]
So q1 × q2 = C × C = C^2 = ( [A] × [T])^2 = [A]^2 × [T]^2
======== ============
4 x Pi = constant , so it has no dimension
=====

F =mass × acceleration = kg × meter/s^2 = [M] × [LT^-2]
==========
r^2 = meter ^2 = [L]^2
=====

Now putting dimesions in the relation (ii) gives

€ = ( [A]^2 × [T]^2 ) / ( [M] × [LT^-2] ) × ( [L]^2 )
= ( [A]^2 × [T]^2 ) / ( [ M L^3 T^-2] )
= [ M^-1 L^-3 T^(2+2) ] [A]^2
= [ M^-1 L^-3 T^4 ] [A]^2